1.

If  \(\begin{bmatrix} x & 3x-y\\[0.3em] 2x+z & 3y-\omega \\[0.3em] \end{bmatrix} ​\)\(=\begin{bmatrix} 3 & 2 \\[0.3em] 4 & 7 \\[0.3em] \end{bmatrix},​​\) find x,y,z,ω.

Answer»

Given two matrices are equal,

   \(\begin{bmatrix} x & 3x-y\\[0.3em] 2x+z & 3y-\omega \\[0.3em] \end{bmatrix} ​\)\(=\begin{bmatrix} 3 & 2 \\[0.3em] 4 & 7 \\[0.3em] \end{bmatrix}\) 

We know that if two matrices are equal then the elements of each matrices are also equal.

∴x = 3 …(1) 

And 3x – y = 2 …(2) 

And 2x + z = 4 …(3) 

3y – ω = 7 …(4) 

Putting the value of x in equation (2),

3 × 3 – y = 2 

⇒ 9 – y = 2 

⇒ y = 9 – 2 

⇒ y = 7

Now, putting the value of y in equation (4),

3×7 – ω = 7 

⇒ 21 – ω = 7 

⇒ ω = 21 – 7 

⇒ ω = 14

Again, Putting the value of x in equation (3),

2 × 3 + z = 4 

⇒ 6 + z = 4 

⇒ z = 4 – 6 

⇒ z = – 2 

∴ x = 3, y = 7, 

z = – 2 and ω = 14



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